Given

${I}_{n}={\int}_{-1}^{1}(1-{x}^{2}{)}^{n}\mathrm{cos}(\theta x)dx,$

${I}_{n}={\int}_{-1}^{1}(1-{x}^{2}{)}^{n}\mathrm{cos}(\theta x)dx,$

Sonia Ayers
2022-07-09
Answered

Given

${I}_{n}={\int}_{-1}^{1}(1-{x}^{2}{)}^{n}\mathrm{cos}(\theta x)dx,$

${I}_{n}={\int}_{-1}^{1}(1-{x}^{2}{)}^{n}\mathrm{cos}(\theta x)dx,$

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Tanner Hamilton

Answered 2022-07-10
Author has **12** answers

Looks like it will be easier to prove it using strong induction on n

Show the base case holds. Assume the induction hypothesis for $k\le n$, ie,

$\begin{array}{}\text{(}\u2020\text{)}& \mathrm{\forall}k\le n:\phantom{\rule{1em}{0ex}}{\theta}^{2k+1}{I}_{k}=k!({P}_{k}(\theta )\mathrm{sin}(\theta )+{Q}_{k}(\theta )\mathrm{cos}(\theta ))\end{array}$

Now, show the inductive step for $n+1$, ie,

$\begin{array}{rl}{\theta}^{2(n+1)+1}{I}_{n+1}& ={\theta}^{2n+1}{\theta}^{2}{I}_{n+1}\\ & \stackrel{(1)}{=}{\theta}^{2n+1}{\textstyle (}(2n+2)(2n+1){I}_{n}-(2n+2)(2n){I}_{n-1}{\textstyle )}\\ & =(2n+2)(2n+1)[{\theta}^{2n+1}{I}_{n}]-(2n+2)(2n){\theta}^{2}[{\theta}^{2(n-1)+1}{I}_{n-1}]\\ & \stackrel{(\u2020)}{=}(2n+2)(2n+1)n!{\textstyle (}{P}_{n}(\theta )\mathrm{sin}\theta +{Q}_{n}(\theta )\mathrm{cos}\theta {\textstyle )}-{\theta}^{2}(2n+2)(2n){\textstyle (}(n-1)!{P}_{n-1}(\theta )\mathrm{sin}\theta +{Q}_{n-1}\mathrm{cos}\theta {\textstyle )}\\ & =2(2n+1)(n+1)!{\textstyle (}{P}_{n}(\theta )\mathrm{sin}\theta +{Q}_{n}(\theta )\mathrm{cos}\theta {\textstyle )}-4(n+1)!{\textstyle (}{\theta}^{2}{P}_{n-1}(\theta )\mathrm{sin}\theta +{\theta}^{2}{Q}_{n-1}\mathrm{cos}\theta {\textstyle )}\\ & =(n+1)!({P}_{n+1}(\theta )\mathrm{sin}\theta +{Q}_{k+1}(\theta )\mathrm{cos}\theta )\end{array}$

where ${P}_{n+1}(\theta ):=2(2n+1){P}_{n}(\theta )-4{\theta}^{2}{P}_{n-1}(\theta )$ and ${Q}_{n+1}(\theta ):=2(2n+1){Q}_{n}(\theta )-4{\theta}^{2}{Q}_{n-1}(\theta )$ and we can finish noting that ${P}_{n+1}$ and ${Q}_{n+1}$ are both polynomials with integer coefficients and degree at most $2n+2=2(n+1)$

Show the base case holds. Assume the induction hypothesis for $k\le n$, ie,

$\begin{array}{}\text{(}\u2020\text{)}& \mathrm{\forall}k\le n:\phantom{\rule{1em}{0ex}}{\theta}^{2k+1}{I}_{k}=k!({P}_{k}(\theta )\mathrm{sin}(\theta )+{Q}_{k}(\theta )\mathrm{cos}(\theta ))\end{array}$

Now, show the inductive step for $n+1$, ie,

$\begin{array}{rl}{\theta}^{2(n+1)+1}{I}_{n+1}& ={\theta}^{2n+1}{\theta}^{2}{I}_{n+1}\\ & \stackrel{(1)}{=}{\theta}^{2n+1}{\textstyle (}(2n+2)(2n+1){I}_{n}-(2n+2)(2n){I}_{n-1}{\textstyle )}\\ & =(2n+2)(2n+1)[{\theta}^{2n+1}{I}_{n}]-(2n+2)(2n){\theta}^{2}[{\theta}^{2(n-1)+1}{I}_{n-1}]\\ & \stackrel{(\u2020)}{=}(2n+2)(2n+1)n!{\textstyle (}{P}_{n}(\theta )\mathrm{sin}\theta +{Q}_{n}(\theta )\mathrm{cos}\theta {\textstyle )}-{\theta}^{2}(2n+2)(2n){\textstyle (}(n-1)!{P}_{n-1}(\theta )\mathrm{sin}\theta +{Q}_{n-1}\mathrm{cos}\theta {\textstyle )}\\ & =2(2n+1)(n+1)!{\textstyle (}{P}_{n}(\theta )\mathrm{sin}\theta +{Q}_{n}(\theta )\mathrm{cos}\theta {\textstyle )}-4(n+1)!{\textstyle (}{\theta}^{2}{P}_{n-1}(\theta )\mathrm{sin}\theta +{\theta}^{2}{Q}_{n-1}\mathrm{cos}\theta {\textstyle )}\\ & =(n+1)!({P}_{n+1}(\theta )\mathrm{sin}\theta +{Q}_{k+1}(\theta )\mathrm{cos}\theta )\end{array}$

where ${P}_{n+1}(\theta ):=2(2n+1){P}_{n}(\theta )-4{\theta}^{2}{P}_{n-1}(\theta )$ and ${Q}_{n+1}(\theta ):=2(2n+1){Q}_{n}(\theta )-4{\theta}^{2}{Q}_{n-1}(\theta )$ and we can finish noting that ${P}_{n+1}$ and ${Q}_{n+1}$ are both polynomials with integer coefficients and degree at most $2n+2=2(n+1)$

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